Walking at `7/9` of his usual speed, a person reaches his office 10 minutes later than the usual time. His usual time in minutes is:

Options

A. 35
B. 27
C. 42
D. 30

Answer

To solve the problem, we need to find the usual time taken by the person to reach his office. Let's break down the solution step by step.

### Step 1: Understand the relationship between speed, time, and distance
We know that:
- Speed = Distance / Time
- If the speed decreases, the time taken increases for the same distance.

### Step 2: Define the variables
Let:
- Usual speed = S
- Usual time = T (in minutes)
- Distance to office = D

From the formula, we can say:
\[ D = S \times T \]

### Step 3: Calculate the new speed and time
When the person walks at \( \frac{7}{9} \) of his usual speed, his new speed becomes:
\[ \text{New speed} = \frac{7}{9} S \]

Since the distance remains the same, we can express the new time taken (T') as:
\[ D = \left(\frac{7}{9} S\right) \times T' \]

### Step 4: Set up the equation for the new time
We know that he reaches 10 minutes late, which means:
\[ T' = T + 10 \]

### Step 5: Substitute the distance in terms of usual time
From the distance equation:
\[ S \times T = \left(\frac{7}{9} S\right) \times (T + 10) \]

### Step 6: Cancel out S (assuming S is not zero)
We can cancel S from both sides:
\[ T = \frac{7}{9} (T + 10) \]

### Step 7: Solve for T
Multiply both sides by 9 to eliminate the fraction:
\[ 9T = 7(T + 10) \]
\[ 9T = 7T + 70 \]
Now, isolate T:
\[ 9T - 7T = 70 \]
\[ 2T = 70 \]
\[ T = 35 \]

### Conclusion
The usual time taken by the person to reach his office is **35 minutes**.